Fatemeh Zahedi HW4 1215240130 STP 598

Problem 1

$y_i$ ~ $Poisson(\lambda)$ $\\P(Y=y|\lambda)=\frac{e^{-\lambda} \, \lambda^{y_i}}{y_i!}$ $\\ P(y_1,y_2,...,y_n;\lambda)={\Pi}\frac{e^{-\lambda} \, \lambda^{y_i}}{y_i!}$ $\\ lnP=ln({\Pi}\frac{e^{-\lambda} \, \lambda^{y_i}}{y_i!})$ $\\ lnP={\Sigma}ln(\frac{e^{-\lambda} \, \lambda^{y_i}}{y_i!})$ $\\ lnP={\Sigma}(ln(\lambda^{y_i})+ln(e^{-\lambda})-ln({y_i!}))$ $\\ lnP={\Sigma}({y_i}ln(\lambda)-\lambda-ln({y_i!}))$ $\\ lnP=-n\lambda+ln(\lambda){\Sigma}{y_i}-{\Sigma}ln({y_i!})$ $\\ \frac{d}{d\lambda}lnP=\frac{d}{d\lambda}(-n\lambda+ln(\lambda){\Sigma}{y_i}-{\Sigma}ln({y_i!}))$ $\\ \frac{d}{d\lambda}lnP=-n+\frac{1}{\lambda}{\Sigma}(y_i)$ $\\ -n+\frac{1}{\lambda}{\Sigma}(y_i)=0$ $\\ {\lambda}=\frac{1}{n}{\Sigma}(y_i)$

Problem 2

$\underset{\omega}min Var(P)=\underset{\omega}{min} \omega'\Sigma\omega$ $\\s.t.$ $\\\omega'1=1$ Using the Lagrange problem then $\\L(\lambda,\omega)=\omega'\Sigma\omega+\lambda(1-\omega'1)$ First order conditions of the Lagrangian problem are $\\ \Sigma\omega-\lambda1=0$ $\\ \omega'1=1$ from the first equation we have $\\ \Sigma\omega = \lambda1$ $\\ \omega = \lambda\Sigma^{-1}1$ Also we have $\\ \omega1=1$ then $\\ \lambda1'\Sigma^{-1}1=1$ $\\ \lambda=\frac{1}{1'\Sigma^{-1}1}$ therefore, the weights of the global-minimum variance portfolio is $\\ \omega^{*}=\frac{1}{1'\Sigma^{-1}1}\Sigma^{-1}1$ And finally the variance of the global-minimum variance portfolio is $\\ Var(\omega^{*}) =\omega^{*}\Sigma\omega^{*}$ $\\ Var(\omega^{*}) = \frac{1}{(1'\Sigma^{-1}1)^{2}}1'\Sigma^{-1}\Sigma\Sigma^{-1}1=\frac{1}{1'\Sigma^{-1}1}$

Problem 3

Problem 4

Comparing mean squared error of these three methods, ridge performed better than the others.